Finding Sine And Tangent Given Cosine And Interval

Hey guys! Let's dive into a cool math problem where we need to figure out the values of sine and tangent, but we're given the cosine and a specific interval. It might sound a bit tricky at first, but trust me, we'll break it down step by step so it’s super clear. So, let's get started and make math a little less mysterious and a lot more fun!

Understanding the Problem

So, the problem states that we have $\cos(\theta) = \frac{\sqrt{2}}{2}$, and we know that $\theta$ lies between $\frac{3\pi}{2}$ and $2\pi$. What we need to find are the exact values of $\sin(\theta)$ and $\tan(\theta)$. Now, the crucial part here is understanding what this interval $\frac{3\pi}{2} < \theta < 2\pi$ tells us. This range corresponds to the fourth quadrant on the unit circle. Why is that important? Well, in the fourth quadrant, cosine is positive, sine is negative, and tangent is also negative. Remembering this quadrant information is super helpful because it guides us in determining the signs of our answers.

To really nail this down, let's visualize the unit circle. Imagine a circle with a radius of 1 centered at the origin of a coordinate plane. The angle $\theta$ starts from the positive x-axis and rotates counterclockwise. When $\theta$ is between $\frac{3\pi}{2}$ (which is 270 degrees) and $2\pi$ (which is 360 degrees), our point on the circle lands in the bottom-right quadrant – the fourth quadrant. So, knowing our location immediately tells us about the signs of sine and cosine.

Also, keep in mind the fundamental trigonometric identity: $\sin^2(\theta) + \cos^2(\theta) = 1$. This is like our secret weapon in solving these problems. We know the value of $\cos(\theta)$, and we need to find $\sin(\theta)$, so this identity is exactly what we need. We’ll plug in the value of cosine and then solve for sine. But hold on! Don’t forget that sine will be negative in the fourth quadrant. Keeping track of these sign conventions is vital to avoid mistakes and get the right answer. Alright, let's put this all together and move on to the next step.

Finding $\sin(\theta)$

Alright, let's get our hands dirty and find the value of $\sin(\theta)$. Remember our fundamental trigonometric identity? It’s $\sin^2(\theta) + \cos^2(\theta) = 1$. This is the key to unlocking our solution. We already know that $\cos(\theta) = \frac{\sqrt{2}}{2}$, so we can plug this value into the identity. This gives us $\sin^2(\theta) + \left(\frac{\sqrt{2}}{2}\right)^2 = 1$.

Let's simplify this. Squaring $\frac{\sqrt{2}}{2}$ gives us $\frac{2}{4}$, which simplifies further to $\frac{1}{2}$. So our equation now looks like $\sin^2(\theta) + \frac{1}{2} = 1$. Next, we want to isolate $\sin^2(\theta)$, so we subtract $\frac{1}{2}$ from both sides of the equation. This gives us $\sin^2(\theta) = 1 - \frac{1}{2}$, which simplifies to $\sin^2(\theta) = \frac{1}{2}$.

Now, to find $\sin(\theta)$, we need to take the square root of both sides. Remember, when you take the square root, you generally get two solutions: a positive and a negative one. So, we have $\sin(\theta) = \pm\sqrt{\frac{1}{2}}$. This simplifies to $\sin(\theta) = \pm\frac{1}{\sqrt{2}}$. But wait! We're not done yet. We need to rationalize the denominator, which means getting rid of the square root in the bottom. To do this, we multiply both the numerator and the denominator by $\sqrt{2}$. This gives us $\sin(\theta) = \pm\frac{\sqrt{2}}{2}$.

Okay, now we have two possible values for $\sin(\theta)$: $\frac{\sqrt{2}}{2}$ and $-\frac{\sqrt{2}}{2}$. This is where our knowledge of the quadrant comes in clutch. We know that $\theta$ is in the fourth quadrant, where sine is negative. So, we can confidently discard the positive value. Therefore, $\sin(\theta) = -\frac{\sqrt{2}}{2}$. See how understanding the quadrant helps us narrow down the possibilities? Now that we’ve found sine, let’s tackle tangent.

Finding $\tan(\theta)$

Alright, let’s move on to finding $\tan(\theta)$. We’ve already figured out that $\sin(\theta) = -\frac{\sqrt{2}}{2}$ and we were given that $\cos(\theta) = \frac{\sqrt{2}}{2}$. So, how do we connect these to find tangent? The key is the definition of the tangent function: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. This formula is like the secret sauce that brings sine and cosine together to give us tangent.

Now, let's plug in the values we know. We have $\tan(\theta) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$. Woah, looks a bit complex, right? But don’t worry, it's actually quite simple. We're dividing a fraction by another fraction, and in this case, they're almost identical except for the negative sign. When you divide a number by itself, you get 1. So, dividing $\frac{\sqrt{2}}{2}$ by $\frac{\sqrt{2}}{2}$ gives us 1.

But we have that negative sign in the numerator, so we need to carry that along. Therefore, $\tan(\theta) = -1$. And that's it! We’ve found the value of $\tan(\theta)$. It’s super neat how the relationship between sine, cosine, and tangent simplifies the calculation. Just remember, tangent is the ratio of sine to cosine, and keeping track of the signs is crucial.

So, to recap, we used the definition of tangent and the values of sine and cosine to quickly find the tangent. This highlights the importance of knowing your trigonometric identities and how they relate to each other. Now, let's wrap up and summarize our findings.

Conclusion

Alright, guys, we made it! We’ve successfully navigated through the problem of finding $\sin(\theta)$ and $\tan(\theta)$ given $\cos(\theta) = \frac{\sqrt{2}}{2}$ and the interval $\frac{3\pi}{2} < \theta < 2\pi$. Let's take a moment to recap our journey and solidify what we’ve learned.

First, we started by understanding the problem and recognizing the importance of the given interval. Knowing that $\theta$ lies in the fourth quadrant was crucial because it told us that sine is negative and cosine is positive. This knowledge acted as our compass, guiding us in the right direction.

Next, we tackled finding $\sin(\theta)$. We used the fundamental trigonometric identity $\sin^2(\theta) + \cos^2(\theta) = 1$ as our primary tool. By plugging in the value of $\cos(\theta)$ and solving for $\sin(\theta)$, we found two possible values: $\pm\frac{\sqrt{2}}{2}$. But remember, our quadrant knowledge came to the rescue, and we correctly identified that $\sin(\theta)$ must be negative in the fourth quadrant. So, we concluded that $\sin(\theta) = -\frac{\sqrt{2}}{2}$.

Finally, we moved on to finding $\tan(\theta)$. Here, we used the definition of tangent: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. By plugging in the values we found for sine and were given for cosine, we got $\tan(\theta) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$. This simplified beautifully to $\tan(\theta) = -1$.

So, to summarize, we found that $\sin(\theta) = -\frac{\sqrt{2}}{2}$ and $\tan(\theta) = -1$. The key takeaways from this problem are the importance of knowing your trigonometric identities, understanding the unit circle, and recognizing how the quadrant affects the signs of trigonometric functions. These tools are super helpful not just for this problem, but for a wide range of trig problems. Keep practicing, and you’ll become a trig whiz in no time! Keep up the great work, and remember, math can be fun when you break it down step by step. You got this!